Open access peer-reviewed chapter - ONLINE FIRST

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance in K-Shell in a Classical Way

By Edward Henry Jimenez

Submitted: September 9th 2020Reviewed: December 7th 2020Published: January 15th 2021

DOI: 10.5772/intechopen.95405

Downloaded: 44

Abstract

First, the liquid drop model assumes a priori; to the atomic nucleus composed of protons and neutrons, as an incompressible nuclear fluid that should comply with the Navier–Stokes 3D equations (N-S3D). Conjecture, not yet proven, however, this model has successfully predicted the binding energy of the nuclei. Second, the calculation of nuclear pressure p0∈1.42,1.94]1032Pa and average viscosity η=1.71×1024fm2/s, as a function of the nuclear decay constant k=p02η=1T1/2, not only complements the information from the National Nuclear Data Center, but also presents an analytical solution of (N- S3D). Third, the solution of (N-S3D) is a Fermi Dirac generalized probability function Pxyzt=11+ep02ηt−μx2+y2+z21/2, Fourth, the parameter μ has a correspondence with the Yukawa potential coefficient μ=αm=1/r, Fifth, using low energy X-rays we visualize and measure parameters of the nuclear surface (proton radio) giving rise to the femtoscope. Finally, we obtain that the pressure of the proton is 8.14 times greater than the pressure of the neutron, and 1000 times greater than the pressure of the atomic nucleus. Analyzed data were isotopes: 9≤Z≤92 and 9≤N≤200.

Keywords

  • femtoscope
  • Navier Stokes 3D
  • nuclear viscosity
  • minimum entropy

1. Introduction

Neutron stars are among the densest known objects in the universe, withstanding pressures of the order of 1034Pa.However, it turns out that protons [1], the fundamental particles that make up most of the visible matter in the universe, contain pressures 10 times greater, [2, 3] 1035Pa.This has been verified from two perspectives at the Jefferson Laboratory, MIT [1, 2, 3, 4, 5]. High-energy physics continue to guide the study of the mechanical properties of the subatomic world.

Figure 1.

Obtaining nuclear viscosity and nuclear pressure from the speed of the neutron particles in the disintegration of chemical element. Figure 1a. indicates that the BE/A ratio is proportional to the nuclear pressure, p0 represented in Figure 1b. Figure 1c is the graph of the viscosity in equilibrium and out of equilibrium, the viscosity in equilibrium is greater than the viscosity at the moment of nuclear decay. Figure 1d is the average half-time of each isotope, 9≤Z≤92 and 9≤N≤200.

Viscosity is a characteristic physical property of all fluids, which emerges from collisions between fluid particles moving at different speeds, causing resistance to their movement. When a fluid is forced to move by a closed surface, similar to the atomic nucleus, the particles that make up the fluid move slower in the center and faster on the walls of the sphere. Therefore, a shear stress (such as a pressure difference) is necessary to overcome the friction resistance between the layers of the nuclear fluid. For the same radial velocity profile, the required tension is proportional to the viscosity of the nuclear fluid or its composition given by ZN[6, 7, 8].

Radioactive decay is a stochastic process, at the level of individual atoms. According to quantum theory, it is impossible to predict when a particular atom will decay, regardless of how long the atom has existed. However, for a collection of atoms of the same type [8], the expected decay rate is characterized in terms of its decay constant k=1T1/2. The half-lives of radioactive atoms do not have a known upper limit, since it covers a time range of more than 55 orders of magnitude, from almost instantaneous to much longer than the age of the universe [8, 9].

Other characteristics of the proton such as its size have been studied in many institutes such as Max Plank, where it has been measured with high precision ranges rp=0.8418467fm,providing new research methods [10, 11].

The atomic nucleus is an incompressible fluid, justified by the formula of the nuclear radius, R=1.2A13, where it is evident that the volume of the atomic nucleus changes linearly with A=Z+N, giving a density constant [11]. All incompressible fluid and especially the atomic nucleus comply with the Navier Stokes equations. We present a rigorous demonstration on the incomprehensibility of the atomic nucleus, which allows to write explicitly the form of the nuclear force FN=gμ28πA1P1Pr, which facilitates the understanding of nuclear decay.

The Navier Stokes equations are a problem of the millennium [12, 13], that has not been resolved yet in a generalized manner. We present a solution that logically meets all the requirements established by the Clay Foundation [14, 15]. This solution coherently explains the incompressible nuclear fluid and allows calculations of the nuclear viscosity and nuclear pressure [1, 2].

The alpha particle is one of the most stable. Therefore it is believed that it can exist as such in the heavy core structure. The kinetic energy typical of the alpha particles resulting from the decay is in the order of 5 MeV.

For our demonstrations, we will use strictly the scheme presented by Fefferman inhttp://www.claymath.org/millennium-problems[13, 14], where six demonstrations are required to accept as valid a solution to the Navier–Stokes 3D Equation [16, 17, 18]. An understanding of the mechanics of the atomic nucleus cannot do without fluid equations.

2. Model

The velocity defined as u=2νPP,with a radius noted as r=x2+y2+z21/2where Pxyztis the logistic probability function Pxyzt=11+ektμr,and the expected value Err0<Cexist. The term Pis defined in xyzR3t0,where constants k>0,μ>0and Pxyxtis the general solution of the Navier–Stokes 3D equation, which has to satisfy the conditions (1) and (2), allowing us to analyze the dynamics of an incompressible fluid [12, 13, 14].

ut+u.u=ν2upρ0xyzR3t0E1

With, uR3an known velocity vector, ρ0constant density of fluid, ηdynamic viscosity, νcinematic viscosity, and pressure p=p0Pin xyzR3t0.

Where velocity and pressure are depending of rand t. We will write the condition of incompressibility.

.u=0xyzR3t0E2

The initial conditions of fluid movement u0xyz, are determined for t=0. Where speed u0must be Cdivergence-free vector.

uxyz0=u0xyzxyzR3E3

For physically reasonable solutions, we make sure uxyztdoes not grow large as r.We will restrict attention to initial conditions u0that satisfy.

xαu0CαK1+rKonR3foranyαandKE4

The Clay Institute accepts a physically reasonable solution of (1), (2) and (3), only if it satisfies:

p,uCR3×0)E5

and the finite energy condition [14, 15, 16].

R3uxyzt2dxdydzCforallt0bounded energy.E6

The problems of Mathematical Physics are solved by the Nature, guiding the understanding, the scope, the limitations and the complementary theories. These guidelines of this research were: the probabilistic elements of Quantum Mechanics, the De Broglie equation and the Heisenberg Uncertainty principle.

2.1 Definitions

Nuclear reaction velocity coefficient.

We will use an equation analogous to concentration equation of Physical Chemistry C=C0ekt, where k=p02η,is velocity coefficient, p0is the initial pressure of our fluid, ηthe dynamic viscosity and C0the initial concentration of energetic fluid molecules.

It is evident that, in equilibrium state we can write μr=kt, however, the Navier–Stokes equation precisely measures the behavior of the fluids out of equilibrium, so that: μrkt.

Fortunately, there is a single solution for out-of-equilibrium fluids, using the fixed-point theorem for implicit functions, 11+ektμr=2μr,the proof is proved in Theorem 1.

Attenuation coefficient.

We will use the known attenuation formula of an incident flux I0,for which I=I0eμr. Where, I0initial flux and μattenuation coefficient of energetic molecules that enter into interaction and/or resonance with the target molecules, transmitting or capturing the maximum amount of energy [5].

Dimensional analysis and fluid elements.

We will define the respective dimensional units of each one of variables and physical constants that appear in the solution of the Navier–Stokes 3D equation [12, 13, 14, 16].

Nuclear decay Nt=N0ekt.Where k=1T1/2[1/s], the velocity coefficient, and T1/2ground state half-life.

Kinematic viscosity ν=ηρ0,m2s.

Dynamic viscosity η,[pa.s], where pa represents pascal pressure unit.

Initial Pressure of out of equilibrium. p0,[pa].

Fluid density ρ0,kgm3, where kgis kilogram and m3cubic meters.

Logistic probability function, Pxyzt=11+ektμr,it is a real number 0P1.

Equilibrium condition, r=kμt=p02ρ0νμt=uet,[m].

Fluid velocity in equilibrium, ue,m/s.Protons, neutrons and alpha particles are the elements of the fluid.

Fluid field velocity out of equilibrium, u=2νμ1Pr.[m/s]. All nuclear decay is a process out of nuclear equilibrium.

Position, r=x2+y2+z21/2,[m].

Attenuation coefficient, μ,[1/m].

Growth coefficient, k=p02ρ0ν=p02η,[1/s].

Concentration C=C01PP.

Theorem 1The velocity of the fluid is given byu=2νPP,wherePxyztis the logistic probability functionPxyxt=11+ektμr,andppressure such thatp=p0P,both defined onxyzR3t0.The functionPis the general solution of the Navier Stokes equations, which satisfies conditions (1) and (2).

Proof. To verify condition (2), .u=0,we must calculate the gradients and laplacians of the radius. r=xryrzr,and 2r=.r=y2+z2+x2+z2+x2+y2x2+y2+z23/2=2r.

.u=2ν.PP=2νμ1PrE7

Replacing the respective values for the terms: 2rand r2in the Eq. (7).

.u=2νμ1Pr=2νμ1Pr=2νμμPP2r2+1P2rE8

Where the gradient modulus of P=μPP2r, has the form P2=μ2PP22r2=μ2PP22.

.u=2νμ1PμP+2r=0E9

Simplifying for 1P0, we obtain the main result of this paper, which represents a fixed point of an implicit function ftrwhere ftr=P2μr=0. In Nuclear Physics, r0<r<1.2A1/3.

P=11+ektμx2+y2+z21/2=2μx2+y2+z21/2xyzR3t0E10

Eq. (10) has a solution according to the fixed-point theorem of an implicit function, and it is a solution to the Navier Stokes stationary equations, which are summarized in: 2P=2μ21r=0.Furthermore, it is the typical solution of the Laplace equation for the pressure of the fluid 2p=p02P=0. Kerson Huang (1987).

To this point, we need to verify that Eq. (10) is also a solution of requirement (1), ut+u.u=ν2upρ0. We will do the equivalence u=θafter we replace in Eq. (1). Taking into account that θ=2νlnP,and that θis irrotational, ×θ=0, we have: u.u=θ.θ=12θ.θθ××θ=12θ.θ,and 2u=.u××u=.θ××θ=2θ.Simplifying terms in order to replace these results in Eq. (1) we obtain

u.u=12θ.θ=2ν2P2P22u=.u=2θ=0=2νP2P22PP=0

The explicit form of velocity is u=2μν1Pr.Next, we need the partial derivative ut

ut=2μνkP1Pr,pρ0=μp0ρ0P1Pr.

After replacing the last four results u.u,2u, utand pρ0in Eq. (1) we obtain (11).

2μνkP1Pr=2ν2P2P2μp0ρ0P1Pr.E11

The Eq. (11) is equivalent to Eq. (1). After obtaining the term P2P2from the incompressibility equation 2θ=2νP2P2+2PP=0and replacing in Eq. (11).

2μνkP1Pr=2ν22PPμp0ρ0P1Pr.E12

Eq. (10) simultaneously fulfills requirements (1) expressed by Eq. (12) and requirement (2) expressed by Eq. (7), for a constant k=p02ρ0ν=p02η.Moreover, according to Eq. (10), the probability P=2μrwhich allows the Laplace equation to be satisfied: 2P=2μ21r=0.In other words, the Navier–Stokes 3D equation system is solved.▪

Implicit Function.

An implicit function defined as (10), ftr=11+ektμr2μr=0has a fixed point trof R=tr0<atb0<r<+, where mand Mare constants, such as: mM.Knowing that the partial derivative exists: rftr=νP1P+2μr2we can assume that: 0<mrftrM.If, in addition, for each continuous function φin abthe composite function gt=ftφtis continuous in ab,then there is one and only one function: r=φtcontinuous in ab, such that ftφt=0for all tin ab.

Theorem 2An implicit function defined as(10)ftr=11+ektμr2μr=0has a fixed pointtrofR=tr0<atb0<r<+. In this way, the requirements (1) and (2) are fulfilled.

Proof. Let Cbe the linear space of continuous functions in ab, and define an operator T:CCby the equation:

t=φt1Mftφt.

Then we prove that Tis a contraction operator, so it has a unique fixed point r=φtin C. Let us construct the following distance.

tt=φtψtftφtftψtM.

Using the mean value theorem for derivation, we have

ftφtftψt=ϕftztφtψt.

Where ϕtis situated between φtand ψt. Therefore, the distance equation can be written as:

tt=φtψt1ϕftztM

Using the hypothesis 0<mrftrMwe arrive at the following result:

01ϕftϕtM1mM,

with which we can write the following inequality:

tt=φtψt1mMαφψ.E13

Where α=1mM. Since 0<mM,we have 0α<1. The above inequality is valid for all tof ab. Where Tis a contraction operator and the proof is complete, since for every contraction operator T:CCthere exists one and only one continuous function φin C,such that Tφ=φ. Using Eq. (10), which represents the fundamental solution of the Navier–Stokes 3D equation, we verify Eq. (2), which represents the second of the six requirements of an acceptable solution.▪

Proposition 3Requirement (3). The initial velocity can be obtained from:uxyz0=2νPP,where each of the componentsux,uyanduzare infinitely derivable.

uxyz0=u0xyz=2νμ1P0xryrzrxyzR3P0=11+eμr0E14

Proof. Taking the partial derivatives of xnxr,ynxrand znxr.

xnxr=nxn11r+xxn1rynxr=nyn11r+yyn1rznxr=nzn11r+zzn1rE15

Recalling the derivatives of special functions (Legendre), it is verified that there exists the derivative C.

xn1r=1nn!x2+y2+z2n+12Pnxx2+y2+z21/2yn1r=1nn!x2+y2+z2n+12Pnyx2+y2+z21/2zn1r=1nn!x2+y2+z2n+12Pnzx2+y2+z21/2E16

Physically, this solution is valid for the initial velocity, indicated by Eq. (4), where the components of the initial velocity are infinitely differentiable, and make it possible to guarantee that the velocity of the fluid is zero when r[6, 7, 8, 9].

Proposition 4Requirement (4). Using the initial velocity of a moving fluid given byuxyz0=u0xyz=2νμ1P0xryrzr, it is evident that

xαu0CαK1+rKonR3foranyα andK

Proof. Using the initial velocity of a moving fluid given by u0xyz=2νμ1P0xryrzr, we can find each of the components: xαux0,yαuy0and zαuz0.

xαxr2=αxα11r+xxα1rαxα11r+xxα1r

For the three components x,y,zthe results of the partial derivatives are as follows:

xαxr2=α2xα11r2+2αxxα11rxα1r+x2xα1r2yαyr2=α2yα11r2+2αyyα11ryα1r+y2yα1r2zαzr2=α2zα11r2+2αzzα11rzα1r+z2zα1r2E17

Replacing Eq. (17) with the explanatory form of the Legendre polynomials, for the following terms xα11rand xα1r.

xα1r=1αα!x2+y2+z2α+12Pαxx2+y2+z21/2xα11r=1α1α1!x2+y2+z2α2Pα1xx2+y2+z21/2E18

Also, knowing that for each α0, the maximum value of Pα1=1.We can write the following inequality

x2xα1r2x2α!2r2α+12αxxα11rxα1r2α!α1!12α1r2α1α2xα11r2α2α1!2r2αE19

Grouping terms for xαxr2,yαyr2and zαzr2we have the next expressions.

xαxr2r2αx2α!2r22xα!2r+α2α1!2yαyr2r2αy2α!2r22yα!2r+α2α1!2zαzr2r2αz2α!2r22zα!2r+α2α1!2E20

The module of xαu0is given by xαu0=xαxr2+yαyr2+zαzr21/2.Simplifying and placing the terms of Eq. (20) we have

xαu0r2α3α!2+α2α1!22x+y+zα!2r

Taking into consideration that xr1,yr1,zr1the last term xαu0can be easily written that.

xαu02α!2r2α2+xr+yr+zrxαu010α!2r2α

It is verified that there exists Cα=10α!2such that if r0,then xαu00.Thus, we proved requirement (4).▪

According to Mathematics, and giving an integral physical structure to the study, we need to prove that there are the spatial and temporal derivatives of the velocity and pressure components, satisfying the requirement (5).

Proposition 5Requirement (5). The velocity can be obtained from:uxyzt=2νPPand each of the componentsux,uyanduzare infinitely derivable.

uxyzt=2ν2xr2yr2zr2xyzR3Pxyzt=11+ep02ηtμr=2μrE21

Proof. Taking partial derivatives for xnxr2,ynxr2and znxr2.

xnxr2=nxn11r2+xxn1r2ynxr2=nyn11r2+yyn1r2znxr2=nzn11r2+zzn1r2E22

Recalling the derivatives of special functions, it is verified that the derivative Cexists. These derivatives appear as a function of the Legendre polynomials Pn..

xn1r2=1nn!x2+y2+z2n+1Pnxx2+y2+z2yn1r2=1nn!x2+y2+z2n+1Pnyx2+y2+z2zn1r2=1nn!x2+y2+z2n+1Pnzx2+y2+z2E23

There are the spatial derivatives nand the time derivative which is similar to Eq. (25).▪

Proposition 6Requirement (5). The pressure is totally defined by the equivalencepxyzt=p0Pxyztand is infinitely differentiable in each of its components.

pxyzt=p0PxyztxyzR3E24

Proof. Taking partial derivatives for xn1r,yn1rand zn1r, recalling the derivatives of special functions of Eq. (16), it is shown that the derivative C. We only have to find the time derivatives: tnp0P=p0tnP. Using Eq. (21) for P, we have.

tP=kP1Pt2P=k212PP1Pt3P=k316P+6P2P1Pt4P=k4114P+36P224P3P1Pt5P=k5130P+150P2240P3+120P4P1PtnP=ttn1PE25

It is always possible to find the derivative tnPas a function of the previous derivative, since the resulting polynomial of each derivative n1is of degree n.▪

Proposition 7Requirement (6). The energy must be limited in a defined volume and fundamentally it must converge at any time, such thatt0.

R3uxyzt2dxdydzCforallt0bounded energy.

Proof. We will use the explicit form of velocity given in Eq. (21)uxyzt=2νμ1Pr,to obtain the vector module: u2=4ν2μ21P2. Rewriting Eq. (21), and applying a change of variable in: dxdydx=4πr2dr.

R3uxyzt2dxdydz=16πν2μ2r0r21P2drE26

Making another change of variable dP=μP1Pdr. Using (10), replacing r2=2μP2we have

R3uxyzt2dxdydz=16πν2μ2P0P2μP21P2dPμP1P=64πν2μP0P1PP3dPE27

Where radius r,when t0,we have limrP=limr11+expktexpμr=P=1.Moreover, physically if rr00then t0we have limr0P=limr011+expktexpμr=P0=12.Here, a probability 12represents maximum entropy.

R3uxyzt2dxdydz=64πν2μ1/211PP3dP=64πν2μ2P12P21/21R3u2dxdydz32πν2μforallt0E28

In this way the value of the constant Cis C=32πν2μ.Verifying the proposition (6) completely. In general, Eq. (10) can be written ftr+r0=11+ektμr+r02μμr+r0=0and in this way discontinuities are avoided when r0, but this problem does not occur since in the atomic nucleus r0<r<1.2A1/3is satisfied.▪

Lemma 8The irrotational field represented by the logistic probability functionPxyztassociated with the velocityu=2νPP, can produce vortices, due to the stochastic behavior of the physical variablesp0,η,μ.These stochastic variations are in orders lower than the minimum experimental value.

Proof. The implicit function representing the solution of the Navier–Stokes 3D equation, 11+eμrkμt=2μrdepends on the values of initial pressure p0, viscosity νand attenuation coefficient μ. Due to Heisenberg uncertainty principle, these parameters have a variation when we measure and use them, as is the case of the estimate of ξ=rϕtkμ. The function ϕtkμ=kμt,expressly incorporates these results, when <ξ<+. The physical and mathematical realities are mutually conditioned and allow for these surprising results. For a definite tthere exist infinities xyzthat hold the relationship r=x2+y2+z21/2. Moreover, for a definite rthere are infinities tthat respect the fixed-point theorem and create spherical trajectories. When the physical variables k,μvary, even at levels of 1/100 or 1/1000, they remain below the minimum variation of the experimental value. We could try to avoid the existence of trajectories on the spherical surface, for which we must assume that the fluid is at rest or it is stationary, which contradicts the Navier–Stokes 3D equation, where all fluid is in accelerated motion ut0. In short, if there are trajectories in the sphere as long as it is probabilistically possible, this is reduced to showing that the expected value of the radius Err0exists and it is finite.

Derivation of Err0.

The logistic density function for ξwhen Eξ=0and Varξ=σ2is defined by.

hξ=μexpξ1+expξ2,where 1μ=σ3/πis a scale parameter. Given that r=ϕtp0ημ+ξfunction for ris then fr=exprϕ/ττ1+exp(rϕ/τ2,to facilitate the calculations we put ϕ=ϕtkμ=kμt. By definition, the truncated density for rwhen r0is given by frr0=frPr0for r0.Given that the cumulative distribution function for ris given by Fr=11+expktμr,it follows that Pr0=1F0=expϕ)1+expϕ)=11+expϕ). The derivation of Err0then proceeds as follows:

Err0=0μrfrr0dr=1Pr00μrexpktμr1+expktμr2drErr0=1Pr01/212μPμP1PdPP1PE29

We replaced in Eq. (29)dP=μP1Pdrand r2=2μP2of this manner we obtain

Err0=1Pr01/212μPμP1PdPP1P=1Pr02μlog2E30

where we have used the fact that

Pr0=expkt1+expkt
Err0=1Pr02μlog22μlog2forallt0E31

where the last equality follows from an application of the L’Hopital’s rule Pr0=limtexpkt1+expkt=1.▪

3. Results

The main results of applying the Navier Stokes equation to the atomic nucleus, which behaves like an incompressible nuclear fluid, are:

  • The nuclear force and the Navier Stokes force are related.

  • The Cross Sections in Low energy X ray can explain the Golden Ratio σ1σ2that appears in the femtoscope.

  • Navier Stokes Equation and Cross Section in Nuclear Physics.

  • The principle of the femtoscope explains that low energy X-rays produce resonance in layer K.

3.1 The nuclear force and the Navier Stokes force are related

Firstly, we will use the concepts of Classic Mechanics and the formulation of the Yukawa potential, Φr=g4πrA1eμrto find the nuclear force exerted on each nucleon at interior of the atomic core FN=Φr. Also, replace the terms of the potential eμr=1PPand 1r=μ2Pby the respective terms already obtained in Eq. (10).

Φr=gA14πreμr=A18π1PE32

The general form of the Eq. (32), is a function of xyzt.

Φrt=gA14πrektμr=A18π1PrtE33

Secondly, we will obtain the Navier Stokes force equation given by:

dudt=ut+u.u=ut+2ν2P2P2=2μνkP1PrE34

Theorem 9The Nuclear Force and Navier Stokes Force are proportional inside the atomic nucleusFN=CFNS.

Proof.Eq. (34) rigorously demonstrated by theorems and propositions 1 through 8, represent the acceleration of a particle within the atomic nucleus. According to Classical Mechanics the force of Navier Stokes applied to a particle of mass m, would have the form:

FNS=mdudt=2mμνkP1Pr.E35

Proof. The nuclear force on its part would be calculated as follows FN=Φr.

FN=Φr=8πA1P.E36

Replacing the term P=μPP2rof Eq. (7), we obtain

FN=Φr=gμ28πA1P1Pr.E37

It is possible to write nuclear force as a function of speed.

FN=Φr=gμ28πA1P1Pr.

Finally, we can show that the nuclear force and force of Navier Stokes differ at most in a constant C. Equating (35) and (37), we find the value gas a function of the parameters nuclear viscosity ν, attenuation μand growth coefficient of the nuclear reaction k, nucleon mass mand C1.

g=16mπνkμA1CE38

3.2 Cross section and Golden ratio σ1σ2are important elements of the femtoscope

According to NIST and GEANT4 [17], current tabulations of μρrely heavily on theoretical values for the total cross section per atom, σtot, which is related to μρby the following equation:

μρ=σtotuAE39

In (Eq. 39), u=1.6605402×1024gris the atomic mass unit (1/12 of the mass of an atom of the nuclide 12C)4.

The attenuation coefficient, photon interaction cross sections and related quantities are functions of the photon energy. The total cross section can be written as the sum over contributions from the principal photon interactions

σtot=σpe+σcoh+σincoh+σtrip+σph.nE40

Where σpeis the atomic photo effect cross section, σcohand σincohare the coherent (Rayleigh) and the incoherent (Compton) scattering cross sections, respectively, σpairand σtripare the cross sections for electron-positron production in the fields of the nucleus and of the atomic electrons, respectively, and σph.n. is the photonuclear cross section3,4.

We use data of NIST and simulations with GEANT4 for elements Z=11to Z=92and photon energies 1.0721×103MeV to 1.16×101MeV, and have been calculated according to:

μρ=σpe+σcoh+σincoh+σtrip+σph.n/μAE41

The attenuation coefficient μof a low energy electron beam 10,100eVwill essentially have the elastic and inelastic components. It despises Bremsstrahlung emission and Positron annihilation.

σtot=σcoh+σincohE42

3.3 The principle of the femtoscope explains that low energy X-rays produce resonance in K-shell

A resonance region is created in a natural way at the K-shell between the nucleus and the electrons at S-level. The condition for the photons to enter in the resonance region is given by rarn+λ. This resonance region gives us a new way to understand the photoelectric effect. There is experimental evidence of the existence of resonance at K-level due to photoelectric effect, represented by the resonance cross section provided by NIST and calculated with GEANT4 for each atom. In the present work we focus on the resonance effects but not on the origin of resonance region.

The resonance cross section is responsible for large and/or abnormal variations in the absorbed radiation I2I1.

I2I1I2+I12=ρrμAσ2σ1E43

Theorem 10Resonance region. The resonance cross section is produced by interference between the atomic nucleus and the incoming X-rays inside the resonance region, where the boundaries are the surface of the atomic nucleus and K-shell.

The cross section of the atomic nucleus is given by:

σrn=4πrn2=4πA2/3rn2E44

The photon cross section at K-shell depends on the wave length and the shape of the atomic nucleus:

σrn+λ=4πrn+λ2E45

Subtracting the cross sections (13) and (14) we have:

σλ=σrn+λσrn=4π2rnλ+λ2=4π2rpλ+2rnrpλ+λ2E46

The resonance is produced by interactions between the X-rays, the K-shell electrons and the atomic nucleus. The cross sections corresponding to the nucleus is weighted by probability pn and should have a simple dependence of an interference term. This last depends on the proton radius rp or the difference between the nucleus and proton radius (rnrp) according to the following relation

Maxσ2σ1=σ1σ2bσ2σ1=4π2rpλE47

We note that left hand side of Eqs. (46) and (47) should have a factor larger than one due to resonance. The unique factor that holds this requirement is σ1σ2bWhere a,bconstants.

8πr¯λσ2σ1=aσ1σ2bE48

After performing some simulations it is shown that the thermal a represents the dimensionless Rydberg constant a=R=1.0973731568539107.

8000πr¯λσ2σ1=Rσ1σ22.5031E49

We use σ1Z,σ2Zfor represent cross section in resonance, Ris the generalized Rydberg constant for all elements of periodic table, and Zatomic number.

Last equation with a R2=0.9935was demonstrated and constructed using the elements of the solution of the Navier Stokes equations.

σ1σ2=0.0021Z+0.0696

This equation was obtained with a R2=0.9939, and indicates that the ratio of the effective sections fully explain each element of the periodic table.

E=2105Z20.0003Z+0.004

This equation obtained for a R2=0.9996, complements the system of equations that allow to know the simulation values as a function of Zfor σ1Z,σ2Zand EZ, where σ1Z<σ2Z.The Femtoscope equations further demonstrate that energy E=minEis minimum and Shannon entropy S=maxSis maximum in resonance, because in equilibrium σ1Zi=σ2Zi.

The radius of the neutron can be obtained using Eq. (49) in the following way.

r¯=R8000πarN=ANr¯+ZNrp

3.4 Navier Stokes equation and cross section in nuclear physics

The speed needs to be defined as u=2νPP,where Pxyztis the logistic probability function Pxyxt=11+ektμr,r=x2+y2+z21/2defined in xyzR3t0This Pis the general solution of the Navier Stokes 3D equations, which satisfies the conditions (50) and (51), allowing to analyze the dynamics of an incompressible fluid.

ut+u.u=pρ0xyzR3t0E50

Where, uR3an known velocity vector, ρ0constant density of fluid and pressure p=p0PR.

With speed and pressure dependent on rand t.We will write the condition of incompressibility as follows.

.u=0xyzR3t0E51

Theorem 11The velocity of the fluid given by:u=2νPP,wherePxyztis6the logistic probability functionPxyxt=11+ektμx2+y2+z21/2,defined inxyzR3t0is the general solution of the Navier Stokes equations, which satisfies conditions(50)and(51).

Proof. Firstly, we will make the equivalence u=θand replace it in Eq. (50). Taking into account that θis irrotational, ×θ=0, we have.

u.u=θ.θ=12θ.θθ××θ=12θ.θ,

We can write,

θt+12θ.θ=p

It is equivalent to,

θt+12θ.θ=Δpρ0

where Δpis the difference between the actual pressure pand certain reference pressure p0.Now, replacing θ=2νlnP,Navier Stokes equation becomes.

Pt=Δpρ0PE52

The external force is zero, so that there is only a constant force Fdue to the variation of the pressure on a cross section σ. Where σis the total cross section of all events that occurs in the nuclear surface including: scattering, absorption, or transformation to another species.

F=pσ2=p0σ1Δp=pp0=σ1σ21p0=1Pp0E53

putting (52) in (53) we have

Pt=μk1PPE54

In order to verify Eq. (51), .u=0,we need to obtain r=xryrzr,2r=.r=y2+z2+x2+z2+x2+y2x2+y2+z23/2=2r.

.u=2ν.PP=2νμ1PrE55

Replacing the respective values for the terms: 2Pand P2of Eq. (55). The Laplacian of Pcan be written as follows.

2P=μ12PP.r+μPP22r=μ212PPP2r2+μPP22r=μ212PPP2+μPP22rE56

Using gradient P=μPP2r, modulus P2=μ2PP22r2and 2Pin (56).

2PPP2P2=0E57

Replacing Eqs. (55) and (56) in (57) we obtain the main result of the Navier Stokes equations, the solution represents a fixed point of an implicit function ftrwhere ftr=P2μr=0.

P=11+ektμx2+y2+z21/2=2μx2+y2+z21/2xyzR3t0E58

An important result of the Navier Stokes 3D equation, applied to the nuclear fluid of an atom, allows us to advance our understanding of nuclear dynamics and nuclear force.

Corollary 12The nuclear decay constantkis determined by the nuclear pressurep0,and the dynamic nuclear viscosityη,as follows:k=p02η.

Proof. A nuclear decay is a reaction of degree 1, which is explained by the exponential law Nt=N0ekt.Rewriting Eq. (58) as a function of the initial nuclear pressure p0and the nuclear viscosity ηwe can prove that they are related to the nuclear decay constant, in an intrinsic way and allow to explain dynamic nuclear phenomena.

Pxyzt=11+ep02ηtμx2+y2+z21/2=2μx2+y2+z21/2xyzR3t0E59

3.5 Model of nuclear pressure

The work in the nuclear fluid is given by the variation of the energy necessary to form that atomic nucleus, that is, by the excess mass required in the process. Using Eq. (59) we can find the explicit form of nuclear pressure p=p0P=p02μr.

W=ΔEmassexcessZA=0r00π02πpdV=0r00π02πp02μrr2cosθdrdθdϕW=ΔEmassexcessZA=p02μ0r0rdr0πsinθdθ02π=4πp0r02μE60

Applying the mean value theorem of integrals, we know that there is a mean value of the nuclear pressure p¯and the volume of the atomic nucleus V¯of the integral (60)

W=ΔEmassexcessZA=0r00π02πpdV=pζVr0E61

that according to Quantum Mechanics, pζ,Vr0are the observable values. By this restriction we can equalize Eq. (60), (61) and find the value of the initial nuclear pressure as a function of the experimentally measured nuclear pressure: pζ4π3r03=4πp0r02μwith r0=1.2A1/2fm,from where:

pζ=ΔEmassexcessZA43πr03=3p0μr0.E62

For Yukawa’s potential, it is often assumed μ1r0.

3.6 Model of nuclear viscosity

Using the fundamental expression k=p02η=1T1/2obtained from the resolution of the Navier Stokes 3D equations, Eqs. (58) and (59), it is possible to find the average or most probable values of the variables involved pressure, nuclear viscosity and nuclear decay constant as follows: k¯=p¯2η¯.

Explicitly we find the nuclear viscosity as a function of the nuclear decay constant k=k¯and the average value of the nuclear pressure p¯as follows:

η¯=p¯2k¯E63

There is another experimental way of determining nuclear viscosity, through the fuzziness of alpha particles, protons or neutrons ejected in a nuclear decay using the fluid velocity equation u=2νμ1Pr,in which modulo u=2νμ1Pr,replacing the dynamic viscosity ν=ηρ0m2s,we obtain: u=2ηρ0μ1P.

So the second way to find the nuclear viscosity has the form:

η=uρ02μ1P=uρ0rP41PE64

Eq. (64) has a more complicated form and depends on the speed of the fluid particles and the radius from which these particles leave.

3.7 Calculation of nuclear pressure and viscosity

The nuclear decay constant kis determined by the nuclear pressure p0,and the dynamic nuclear viscosity η,as follows: k=p02η.

4. Discussion of results

4.1 Calculation of the pressure relation of proton and neutron

The fine-structure constant, α, has several physical interpretations, we use the most known.

The fine-structure constant, α=1137.03599917435is the ratio of two energies: the energy needed to overcome the electrostatic repulsion between two protons a distance of 2rapart, and (ii) the energy of a single photon of wavelength λ=2πr(or of angular wavelength r; according to Planck relation).

α=e24πε02r/hcλ=e24πε02r2πrhc=e24πε02rrc=12e24πε0c=12αE65

We can find the relationship of energies between the proton and the atomic nucleus. Knowing that the two occupy the same nuclear volume V.This relationship is identical to 1α',since the atomic nucleus interacts with the proton through the electromagnetic field and the nuclear force.

pppN=ppVpNV=2α=274.07E66

Thus, we already know the pressure relation between the proton and the atomic nucleus pppN=2α.

Now we find the relation of pressures between the neutron and the atomic nucleus, which are under the action of the same nuclear force, F.

pnpN=F/π0.841842F/π1.2A1/32=1.2A1/320.841842E67

For the chemical element with maximum nuclear pressure, 2862Niwe have pnpN=1.2621/320.841842=31.830.

If we divide Eq. (67) for Eq. (66) we have.

pppn=pppN/pnpN=274.07/31.830=8.6104

4.2 An action on the nuclear surface produces a reaction in the nuclear volume and vice versa

It is created by the friction between the layers of nucleons.

Theorem 13An action on the nuclear surface produces a reaction in the nuclear volume and vice versa.

Proof. The volume and the nuclear surface are connected through the Gaussian divergence theorem and the Navier Stokes equations.

For an incompressible fluid, whose velocity field uxyzis given, .u=0is fulfilled.

Logically, the integral of this term remains zero, that is:

.udxdydz=0

Writing the Divergence theorem. u.ndS=.udxdydz=0,the first term must be equal to zero, that is:

u.ndS=uncosαdS=0α=π2

The only possible trajectory is circular, because in this case the vector nis perpendicular to the surface of the sphere. In this way the equation of the outer sphere corresponding to the surface is: x2+y2+z2=1.2A1/3.

Within the nuclear fluid there are layers of nucleons that move in spherical trajectories.▪

5. Conclusions

High energy physics is the guide to low energy physics, because, in certain processes such as in the measurement of the internal pressure of protons and neutrons. However, we demonstrate that there is a trend compatibility of the two in the characterization of the atomic nucleus.

The Femtomathematics corresponds to the tools and the logic that allows to calculate the parameters of the order of 1015m.In an indirect way, it is in full correspondence with Femtophysics.

The nuclear viscosity at equilibrium is much larger than the nuclear viscosity at the moment of nuclear decay, which is totally logical, because at equilibrium the atomic nucleus is totally compact in pressure and density, while at the moment protons, neutrons, alpha particles come out of the atomic nucleus from a nuclear decay, indicating that the nuclear viscosity decreased.

Acknowledgments

  1. To the participants of the congress: The 4th International Conference on Materials Sciences and Nanomaterials (ICMSN 2020), Cambridge, United Kingdom during July 08-10, 2020.

  2. To the research directorate (DI) of the Central University of Ecuador, as well as the group of the

GIIP process of the Faculty Chemical Engineering, UCE.

Conflicts of interest

The author declare no conflict of interest.

Download for free

chapter PDF

© 2021 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

How to cite and reference

Link to this chapter Copy to clipboard

Cite this chapter Copy to clipboard

Edward Henry Jimenez (January 15th 2021). Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance in K-Shell in a Classical Way [Online First], IntechOpen, DOI: 10.5772/intechopen.95405. Available from:

chapter statistics

44total chapter downloads

More statistics for editors and authors

Login to your personal dashboard for more detailed statistics on your publications.

Access personal reporting

We are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities.

More About Us